Observe, that \(N_G(Y)=KX=KL\triangleright H\)  with \(L/H\triangleleft KL/H\)  cyclic. If \(K/H\)  is also normal in \(L/H\)  then
all the assumptions hold for \(K\)  in place of \(H\) . So, in the following we assume that either \(K=H\)  or \(K>H\)  and not normal in \(L\) .
It also follows from the above properties
that \(N_{G}(Y)\)  acts on the infinite cyclic group \(X/Y\)  so for any \(g\in N_{G}(Y)\)  we have \(x^g\in xY\)  or \(x^g\in x^{-1}Y\) . Note also that, by (b), \(H\)  trivially acts on \(X/Y\) .
We need show that the permutation character is multiplicity-free in every finite quotient. Let \(N\triangleleft G\)  be any normal subgroup of finite index. It contains \(N\cap X\) , still of finite index in \(X\) . Given this finite index we can pick \(h\)  sufficiently large so that \(N\cap X\supseteq X^{(h)}=\lbrace y^h\mid y\in X\rbrace \)  which is still a normal subgroup of finite index in \(G\) . If \(ZX^{(h)}/X^{(h)}\)  is a Gelfand subgroup of \(G/X^{(h)}\)  then all the more so is \(ZN/N\)  a Gelfand subgroup of \(G/N\) . So it is enough to consider factors by the normal subgroups \(X^{(h)}\) . For ease of notation we use the same symbol for subgroups of the factor group as for subgroups of \(G\) . In other words, instead of being a free Ableian group, \(X\)  is assumed to be isomorphic to the homogeneous Abelian group \(Z_h^n\) . The rest of Hypothesis REF  remains intact.
In particular, \(L/H\)  is cyclic or order \(h\)  and if \(K>H\)  then \(KL/H\)  is dihedral of order \(2h\) . As \(1_K^{G}=(1_K^{KL})^G\) , first we treat the irreducible constituents of \(1_K^{KL}\)  separately.
Claim1:
Suppose \(\varphi \in \operatorname{\rm Irr}(KL)\)  is a linear constituent of \(1_K^{KL}\)  that is \(\varphi _{\downarrow K}=1_K\) .
Then \(\varphi ^G\)  is multiplicity free.
Indeed, as \(G=WKL\) , we may use a special case of Mackey's theorem [1]} for the restriction of the induced character \({\varphi ^G}_{\downarrow W}=(\varphi _{\downarrow K\cap W})^W=1_{K\cap W}^W\)  which is multiplicity free by (e). Hence \(\varphi ^G\)  itself is also multiplicity free.
Let \(\varepsilon \)  be a primitive \(h\) -th root of 1 and \(\eta \)  a linear character of \(L\)  with kernel \(H\)  and \(\eta (x)=\varepsilon \) .
We consider \(1_{H}^{L}=\sum _{s=1}^{h}\eta ^s\) , a sum of linear characters, each a power of \(\eta \) . For any integer \(i\)  and \(g\in H\)  we have \(\eta ^t(gx^i)=\varepsilon ^{ti}\) . Let \(\lambda _t=\eta ^t_{\downarrow X}\in \operatorname{\rm Irr}(X)\)  for \(t=1,\ldots ,h\) .
Claim2:
If for some \(g\in G\) , \(\lambda _t=\lambda _s^g\) , where \(t\ne h,h/2\)  then \(g\in KL\) , so \(s=t\)  or \(s=h-t\)  and \(K>H\) .
Indeed, if \(g\notin KL=N_G(Y)\)  then, by assumption (d), there exists \(y,z\in Y\)  such that \(y^g=x^mz\)  with \(m\in \lbrace \pm 1,\pm 2\rbrace \) . So \(1=\lambda _s(y)=\lambda _s^g(x^m z)=\lambda _t(x^m)=\varepsilon ^{mt}\) , impossible.
Claim3:
If \((\eta ^t)^G\ne (\eta ^s)^G\)  then they share no common irreducible constituent. Of course, if \(K>H\)  and \(s=h-t\)  then \((\eta ^s)^{KL}=(\eta ^t)^{KL}\)  so \((\eta ^t)^G=(\eta ^s)^G\) .
Note that
the irreducible constituents of \((\eta ^t)^G\)  lie above \(\lambda _t\) . So, by Clifford's Theorem[1]}, it is enough to prove that \(\lambda _t\)  and \(\lambda _s\)  are not \(G\) -conjugate, unless \(s=t\)  or \(s=h-t\)  and \(K>H\) . By Claim2, we have to check only \(s,t\in \lbrace h,h/2\rbrace \) . But then \(s=t\)  as \(\operatorname{\rm Ker}(\lambda _h)=G\ne \operatorname{\rm Ker}(\lambda _{h/2})\) .
If \(K=H\)  then, by Claim1, all \((\eta ^s)^G\)  are multiplicity free and, by Claim3, the \((\eta ^s)^G\)  (\(s=1,\ldots ,h\) ) share no common irreducible constituent. Hence \(1_H^G=\sum _{s=1}^h (\eta ^s)^G\)  is multiplicity free, indeed.
So let \(K>H\) . If \(t=h,h/2\)  (the second only for even \(h\) ) then
there are unique extensions \(1_{KL}\)  of \(\eta ^h=1_L\)  and \(\mu \)  of \(\eta ^{h/2}\)  (for \(h\)  even) to \(KL\)  that are trivial on \(K\) . So, using Claim1 again, we conclude that
\(1_{KL}^G\)  and \(\mu ^G\)  are multiplicity free with all constituents lying above \(\lambda _h=1_X\)  and \(\lambda _{h/2}\) , respectively.
Assume now \(t\ne h,h/2\) . Let \(I_t=I_G(\lambda _t)=\lbrace g\in G\mid \lambda _t^g=\lambda _t\rbrace \ge L\)  denote the inertia subgroup of \(\lambda _t\) . By Claim2,
\(I_t\subseteq KL\) . If \(g\in K\setminus H\)  then \(\lambda _t^g=\lambda _{h-t}\ne \lambda _t\)  so \(I_t=L\)  and \((\eta ^s)^G\)  is irreducible as it is induced from the inertia subgroup[1]}.
Now \(1_K^{KL}\)  is a sum of irreducible, degree 2 characters \(\sum _{s=1}^{\lfloor \frac{h-1}{2}\rfloor }(\eta ^s)^{KL}\) , of \(1_{KL}\)  and if \(h\)  is even then of an additional linear character \(\mu \)  extending \(\eta ^{h/2}\) . For \(1\le s<h/2\)  \((\eta ^s)^G\)  are distinct irreducible while \(\mu ^G\)  (for \(h\)  even)
and \(1_{KL}^G\)  are multiplicity free. By Claim3, there are no shared constituents among these, so \(1_K^G\)  is multiplicity free.
If together with (a) and (b), instead of (c), (d) and (e), the alternatives (c'), (d'), (e') hold then we make the following modifications. For Claim1: if \(\varphi \in \operatorname{\rm Irr}(L)\)  is a linear constituent of \(1_H^{L}\)  then \(\varphi ^G\)  is multiplicity free. The proof is the same, using (e'). For Claim2, the above proof still works, using (d'), unless \(a=\pm 3\)  and \(t=h/3,2h/3\) . But even in that case \(s=t\)  or \(s=h-t\) . So, if for some \(g\in G,\ \lambda _t=\lambda _s^g\)  then \(s=t\)  or \(s=h-t\) .
Claim3 holds, that is \((\eta ^t)^G,(\eta ^s)^G\)  share no common irreducible summand unless \(s=t,s=h-t\)  but in these cases they are the same, as \(K>H\)  by (c'). The conclusion of the proof is the same: we use Claim1 (and (c')) to show that for \(t\ne h,h/2\)  \((\eta ^t)^G=((\eta ^t)^{KL})^G\)  are all multiplicity free. Hence, by Claim3 as above, \(1_K^G\)  is multiplicity free.
